The packaging lists a model airplane’s length as 15 in. It also gives the scale as 1:78. What is the length of the actual airpla
1 answer:
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Answer:
see explanation
Step-by-step explanation:
The sum of the 3 angles in a triangle = 180°
Subtract the sum of the given angles from 180 for B
B = 180° - (90 + 40)° = 180° - 130° = 150°
--------------------------------------------------------------
sin40° = =
Multiply both sides by 21
21 × sin40° = a, thus
a ≈ 13.5 ( to the nearest tenth )
--------------------------------------------------------------
cos40° = =
Multiply both sides by 21
21 × cos40° = b, thus
b ≈ 16.1 ( to the nearest tenth )
Given : supply voltage = 120V;60 Hz
Inductive Reactance Ω
The inductive reactance is given by the formula, , where L is inductance and
is given by formula,
Apply this, we can say that,
Proof of the Law of Sines
The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)
a
sin A
=
b
sin B
=
c
sin C
For more see Law of Sines.
Acute triangles
Draw the altitude h from the vertex A of the triangle
From the definition of the sine function
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Repeat the above, this time with the altitude drawn from point B
Using a similar method it can be shown that in this case
c
sin C
=
a
sin A
Combining (4) and (5) :
a
sin A
=
b
sin B
=
c
sin C
- Q.E.D
Obtuse Triangles
The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.
First the interior altitude. This is the same as the proof for acute triangles above.
Draw the altitude h from the vertex A of the triangle
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Draw the second altitude h from B. This requires extending the side b:
The angles BAC and BAK are supplementary, so the sine of both are the same.
(see Supplementary angles trig identities)
Angle A is BAC, so
sin A =
h
c
or
h = c sin A
In the larger triangle CBK
sin C =
h
a
or
h = a sin C
From (6) and (7) since they are both equal to h
c sin A = a sin C
Dividing through by sinA then sinC:
a
sin A
=
c
sin C
Combining (4) and (9):
a
sin A
=
b
sin B
=
c
sin C
